# Converse of Spectral Theorem for Compact Self-Adjoint Operators

If I have a bounded linear operator $A$ on a Hilbert space $H$ whose eigenvectors form an orthonormal basis for $H$ and whose corresponding eigenvalues go to $0$ then is $A$ compact and self-adjoint?

I ask because I want to prove that $A$ defined on an orthonormal basis $\{e_k\}$ as $Ae_k=e_k/(k^2+1)$ is compact and self-adjoint. I know that it is, but I'm just wondering if appealing to the spectral theorem is valid. Thanks!

• Yes, it's valid. In particular, we can define $A$ as the limit of finite-rank operators (with respect to the operator norm). $A$ inherits self-adjointness from the sequence since $A \mapsto A^*$ is operator-norm continuous. – Omnomnomnom May 22 '17 at 3:20
• Thanks! Just to be clear, this holds for all $A$ with eigenvectors that form an onb and eigenvalues that go to 0 and not just for the explicit operator $A$ I gave, correct? – Robert B. Marshall May 22 '17 at 3:37
• Yep!${}{}{}{}{}{}{}{}{}$ – Omnomnomnom May 22 '17 at 3:39

Yes, $A$ is compact. By considering truncations of $A$ that are zero on $e_k$ for $k\geq n$, you can write $A$ as a limit of finite-rank operators.